Problem 03 - Bernoulli's Energy Theorem

Discharge
$Q_1 = Q_2 = Q = 0.03 \, \text{ m}^3\text{/s}$
 

Head loss
$HL = \dfrac{20}{\gamma} \, \text{ m}$
 

Velocity heads
$\dfrac{v^2}{2g} = \dfrac{8Q^2}{\pi^2gD^4}$

$\dfrac{{v_1}^2}{2g} = \dfrac{8(0.03^2)}{\pi^2(9.81)(0.3^4)} = 0.0092 \, \text{ m}$

$\dfrac{{v_2}^2}{2g} = \dfrac{8(0.03^2)}{\pi^2(9.81)(0.1^4)} = 0.7436 \, \text{ m}$
 

 

Energy equation between 1 and 2
$E_1 - HL = E_2$

$\dfrac{{v_1}^2}{2g} + \dfrac{p_1}{\gamma} + z_1 - HL = \dfrac{{v_2}^2}{2g} + \dfrac{p_2}{\gamma} + z_2$

$0.0092 + \dfrac{200}{\gamma} + 0 - \dfrac{20}{\gamma} = 0.7436 + \dfrac{p_2}{\gamma} + 0$

$\dfrac{p_2}{\gamma} = \dfrac{180}{\gamma} - 0.7344$

$p_2 = 180 - 0.7344\gamma$
 

Part a: The liquid is water:
$p_2 = 180 - 0.7344(9.81)$

$p_2 = 172.79 \, \text{ kPa}$           answer
 

Part b: The liquid is oil (sp gr = 0.80):
$p_2 = 180 - 0.7344(0.80 \times 9.81)$

$p_2 = 174.24 \, \text{ kPa}$           answer
 

Part 3: The liquid is molasses (s = 1.5):
$p_2 = 180 - 0.7344(1.5 \times 9.81)$

$p_2 = 169.19 \, \text{ kPa}$           answer